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\begin{document}

\title{5-伊藤积分的概念 }
%(1.1-1.2) 
%\institute{上海立信会计金融学院}
\author{{\ppr LQW}}
%\renewcommand{\today}{{\ppr \number\year \,年 \number\month \,月 \number\day \,日} }
\date{{\ppr 2023年1月6日} }

\maketitle

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%\begin{frame}[fragile=singleslide]{1.1.1. }
\begin{frame}{内容提要 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
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\begin{enumerate}
\item[5.1.]  例子：布朗运动的黎曼-斯蒂尔杰斯和式的均方收敛
\item[5.5.]  例子：伊藤积分
\item[5.9.]  简单过程
\item[5.12.]  简单过程的伊藤积分
\item[5.18.]  一般过程的伊藤积分
\item[5.E.]  伊藤积分的一些练习题 
\end{enumerate}

\end{frame}

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\begin{frame}{5.1. 布朗运动的一个黎曼-斯蒂尔杰斯和式 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
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\begin{itemize}\itemsep1em

\item  考虑布朗运动的在每个小区间取左端点的黎曼-斯蒂尔杰斯和式 $$S_n = \sum\limits_{i=1}^{n} B_{t_{i-1}}\Delta_i B,$$ 其中

\begin{itemize}
\item  分划 $\tau_n: 0 = t_0 < t_1 < t_2 < \cdots < t_n = t$.
\item  函数 $f$ 在每个小区间的增量 $\Delta f: \Delta_i f = f(t_i) - f(t_{i-1}), \,\, i=1,2,\cdots, n$. 
\item  分划的小区间的长度 $\Delta_i = t_i - t_{i-1},\,\, i=1,2,\cdots, n$.
\item  分划的小区间的最大长度记为 $\text{mesh}(\tau_n) = \underset{1\le i\le n}{\max}\Delta_i$. 
\end{itemize}

\end{itemize}

\end{frame}

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\begin{frame}{5.2. 一个恒等式 }

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\begin{itemize}\itemsep1em
\item  对任意实数 {\footnotesize $b_0,b_1,b_2,\cdots, b_n$}, 成立恒等式  
{\footnotesize 
\begin{eqnarray*}
&& b_0(b_1-b_0) + b_1(b_2-b_1) + \cdots + b_{n-1}(b_n-b_{n-1}) \\ 
&=& \frac{1}{2}b_n^2 -\frac{1}{2}b_0^2 - \frac{1}{2} \left[ (b_1-b_0)^2 + (b_2-b_1)^2 + \cdots + (b_n-b_{n-1})^2 \right].
\end{eqnarray*}
}
\item  证明：记 {\footnotesize $c_0=b_1-b_0,\,\, c_1=b_2-b_1,\,\, \cdots, \,\, c_{n-1}=b_n-b_{n-1}$}, 则上式等价于
{\footnotesize 
\begin{eqnarray*}
2b_0c_0 + 2b_1c_1 + \cdots + 2b_{n-1}c_{n-1}
= b_n^2 - b_0^2 - \left( c_0^2 + c_1^2 + \cdots + c_{n-1}^2 \right).
\end{eqnarray*}
}
将右边第三项移到左边，两边再加上 {\footnotesize $b_0^2+b_1^2+\cdots+b_{n-1}^2$}, 上式等价于
{\footnotesize 
\begin{eqnarray*}
(b_0+c_0)^2 + (b_1+c_1)^2 + \cdots + (b_{n-1}+c_{n-1})^2
= b_n^2 - b_0^2 + \left( b_0^2 + b_1^2 + \cdots + b_{n-1}^2 \right).
\end{eqnarray*}
 }

\end{itemize}

\end{frame}

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\begin{frame}{5.3. 布朗运动的二次变差 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
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\begin{itemize}\itemsep1em

\item  布朗运动的取左端点的黎曼-斯蒂尔杰斯和式 $S_n$ 可以写成 $$S_n = \frac{1}{2}B_t^2 - \frac{1}{2} \sum\limits_{i=1}^{n} (\Delta_i B)^2.$$
\item  使用下述记号，称为布朗运动的关于这个分划的二次变差， $$Q_n(t) := \sum\limits_{i=1}^{n} (\Delta_i B)^2 $$
\item  研究布朗运动的关于给定分划的二次变差的均值与方差，可得
\begin{eqnarray*}
\mathbb{E}[Q_n(t)] &=& t, \\ 
\text{Var}[Q_n(t)] &<& 2t\cdot\text{mesh}(\tau_n)\to 0. 
\end{eqnarray*}

\end{itemize}

\end{frame}

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\begin{frame}{5.4. 均方收敛 }

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\begin{itemize}\itemsep1em

\item  每个 $S_n$ 是一个随机变量，将每条样本路径变成一个实数。
\begin{eqnarray*}
S_n: \Omega &\to& \mathbb{R}, \\
\omega &\mapsto& S_n(\omega) = \sum\limits_{i=1}^{n} B_{t_{i-1}}(\omega) \Delta_i B(\omega). 
\end{eqnarray*}

\item  当分划越来越细的时候，$S_n$ 按照均方意义下收敛到随机变量 
\begin{eqnarray*}
I_t(B): \Omega &\to& \mathbb{R}, \\
\omega &\mapsto& \frac{1}{2}B^2_t(\omega) -\frac{1}{2}t. 
\end{eqnarray*}

\item  {\color{red}在平方可积的随机变量集合中，均方收敛的含义是 
\begin{eqnarray*}
\mathbb{E}[S_n-I_t(B)]^2 \to 0. 
\end{eqnarray*}
}

\end{itemize}

\end{frame}

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\begin{frame}{5.5. 一个伊藤积分的例子 }

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\begin{itemize}\itemsep1em

\item  考察下述等式， $$\int_0^t B_sdB_s = \frac{1}{2} (B_t^2 -t). $$

\item  等式的左边理解为，在每个小区间取左端点的黎曼-斯蒂尔杰斯和式在均方收敛意义下的极限。这样的积分称为伊藤积分。

\item  在均方收敛意义下，左边这个和式的极限是右边。

\end{itemize}

\end{frame}

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\begin{frame}{5.6. 有用的观察 }

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\begin{itemize}\itemsep1em

\item  {\color{red}Integrals with respect to Brownian sample paths, which cannot be defined in the Riemann-Stieltjes sense, can hopefully be defined in the mean square sense. }

\item  The increment $\Delta_i B=B_{t_i} - B_{t_{i-1}}$ on the interval $[t_{i-1},t_i]$ satisfies $\mathbb{E}(\Delta_iB)=0$ and $\mathbb{E}(\Delta_iB)^2=\Delta_i=t_i-t_{i-1}$. 
The mean square limit of $Q_n(t)$ is $t$. These properties suggest that $(\Delta_iB)^2$ is of the order $\Delta_i$. 


\end{itemize}

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\begin{frame}{5.7. 布朗运动的二次变差 }

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\begin{itemize}\itemsep1em

\item  In terms of differentials, we write $$(dB_t)^2 = (B_{t+dt} - B_t)^2 = dt. $$
\item  In terms of integrals, $$\int_0^t (dB_s)^2 = \int_0^t ds =t.$$
\item  The right-hand side is the quadratic variation of Brownian motion on $[0,t]$. 


\end{itemize}

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\begin{frame}{5.8. 伊藤积分与鞅 }

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\begin{itemize}\itemsep1em

\item  继续考虑布朗运动的在每个小区间取左端点的黎曼-斯蒂尔杰斯和式 
$$S_k = \sum\limits_{i=1}^{k} B_{t_{i-1}}(B_{t_i} - B_{t_{i-1}}), \,\, k=1,2,\cdots,n.$$

\item  这是一个鞅变换。

\item  随机过程 $\{S_k\}$ 关于递增的事件域序列 $\{\mathcal{F}_k\}$ 是一个鞅过程，这里 $$\mathcal{F}_k:=\sigma(B_{t_i}, 1\le i\le k).$$

\item  连续时间随机过程 $\{I_t(B)=\frac{1}{2}(B_t^2-t),t\ge 0\}$ 关于布朗运动生成的递增的事件域序列 $\{\mathcal{F}_t=\sigma(B_s,0\le s\le t), t\ge 0\}$ 是一个鞅。

\end{itemize}

\end{frame}

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\begin{frame}{5.9. 简单过程 }

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\begin{itemize}\itemsep1em

\item  {\color{red} 定义：称随机过程 $C=\{C_t, t\in [0,T]\}$ 是一个简单过程，如果存在时间区间的一个分划 
$$\tau_n: 0 = t_0 < t_1 < t_2 < \cdots < t_n = T,$$
以及一列随机变量 $(Z_1,Z_2,\cdots, Z_n)$ 使得
\begin{eqnarray*}
C_t = \left\{ \begin{array}{ll}
Z_1, & t_0\le t < t_1, \\ 
Z_2, & t_1\le t < t_2, \\ 
\cdots & \cdots, \\
Z_{n-1}, & t_{n-2}\le t < t_{n-1}, \\ 
Z_n, & t_{n-1}\le t \le t_n. 
\end{array}\right.
\end{eqnarray*}
}

\end{itemize}

\end{frame}

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\begin{frame}{5.10. 例子：从布朗运动得到简单过程 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
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\begin{itemize}\itemsep1em

\item  取定区间 $[0,T]$ 的一个分划 $\tau_4: 0 = t_0 < t_1 < t_2 < t_3 < t_4 = T$. 
\item  从布朗运动 $\{B_t, t\in [0,T]\}$ 出发，定义一个简单过程
\begin{eqnarray*}
C_t = \left\{ \begin{array}{ll}
B_{t_0}, & t_0\le t < t_1, \\ 
B_{t_1}, & t_1\le t < t_2, \\ 
B_{t_2}, & t_2\le t < t_3, \\ 
B_{t_3}, & t_3\le t \le t_4. 
\end{array}\right.
\end{eqnarray*}

\item  对布朗运动的每条路径 $\omega$, 得到简单过程的一条路径 $\omega'$. 

\end{itemize}

\end{frame}

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\begin{frame}{5.11. 从布朗运动的一条路径得到简单过程的一条路径 }

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\begin{center}
\includegraphics[height=0.7\textheight, width=0.9\textwidth]{simple-process.png}
\end{center}

\end{frame}

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\begin{frame}{5.12. 简单过程的伊藤积分 }

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\begin{itemize}\itemsep1em

\item  {\color{red}定义：简单过程 $\{C_t, t\in [0,T] \}$ 在区间 $[0,T]$ 上的伊藤积分为
\begin{eqnarray*}
\int_0^T C_sdB_s = \sum\limits_{i=1}^{n} C_{t_{i-1}} (B_{t_i} - B_{t_{i-1}}) = \sum\limits_{i=1}^{n} Z_i\Delta_iB. 
\end{eqnarray*}
}
\item  定义：简单过程 $\{C_t, t\in [0,T] \}$ 在区间 $[0,t]$ 上的伊藤积分为
\begin{eqnarray*}
\int_0^t C_sdB_s
 = \left\{ \begin{array}{ll}
Z_1(B_t-B_{t_0}), & t_0\le t < t_1, \\ 
Z_1\Delta_1B + Z_2(B_t - B_{t_1}), & t_1\le t < t_2, \\ 
Z_1\Delta_1B + Z_2\Delta_2B + Z_3(B_t - B_{t_2}), & t_2\le t < t_3, \\ 
\cdots & \cdots \\
Z_1\Delta_1B + \cdots + Z_{n-1}\Delta_{n-1}B + Z_n(B_t - B_{t_{n-1}}), & t_{n-1}\le t \le t_n.  
\end{array}\right.
\end{eqnarray*}

\end{itemize}

\end{frame}

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\begin{frame}{5.13. 前两页的布朗运动与简单过程的伊藤积分 }

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\begin{center}
\includegraphics[height=0.7\textheight, width=0.9\textwidth]{simple-process-ito-integral.png}
\end{center}

\end{frame}

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\begin{frame}{5.14. 伊藤积分与鞅 }

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\begin{itemize}\itemsep1em

\item  {\color{red}定理：设 $\{B_t,\, t\in [0,T]\}$ 是布朗运动，生成递增的事件域 $$\{ \mathcal{F}_t=\sigma(B_u, 0\le u\le t), t\in [0,T]\}. $$
设 $\{C_t,\, t\in [0,T]\}$ 是一个简单随机过程，关于递增的事件域序列 $\{\mathcal{F}_t\}$ 是可预见的。
则伊藤积分 $$I_t(C) = \int_0^t C_sdB_s, \, t\in [0,T]$$
关于递增的事件域序列 $\{\mathcal{F}_t\}$ 是一个鞅。}

\item  证明：验证鞅的定义的几个条件。

\end{itemize}

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\begin{frame}{5.15. 证明从伊藤积分得到鞅 - 1 }

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\begin{itemize}%\itemsep1em

\item  {\color{red}从伊藤积分的等距性质可得 $\mathbb{E}|I_t(C)|<\infty$.} 等距性质如下：
$$\mathbb{E} \left( \int_0^t C_sdB_s \right)^2 = \int_0^t \mathbb{E}(C_s^2)ds, \,\, t\in [0,T]. $$

\item  证明：考虑 $t=T$ 的情形。记 $W_i = Z_i\Delta_iB, \,\, 1\le i\le n$. 则 $$ \int_0^T C_sdB_s = \sum\limits_{i=1}^{n} W_i. $$ 
设 $i>j$, 则 $Z_iZ_j (\Delta_jB)$ 与 $\Delta_iB$ 相互独立，所以
\begin{eqnarray*}
\mathbb{E}(W_iW_j) 
= \mathbb{E} \left[ Z_i Z_j (\Delta_iB) (\Delta_jB) \right] 
&=& \mathbb{E} \left[ Z_iZ_j (\Delta_jB) \right] \mathbb{E} (\Delta_iB) \\
&=& \mathbb{E} \left[ Z_iZ_j(\Delta_iB) \right] \cdot 0 = 0. 
\end{eqnarray*}

\end{itemize}

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\begin{frame}{5.16. 证明从伊藤积分得到鞅 - 2 }

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\begin{itemize}%\itemsep1em

\item  {\color{red}随机过程 $\{I_t(C), t\in [0,T]\}$ 是适应于递增的事件域 $\{\mathcal{F}_t\}$ 的。}

\item  随机变量 $Z_k$ 是 $\mathcal{F}_{t_{k-1}}$-可测的。

\begin{table}[ht]
\centering
\begin{tabular}{|M{2cm}|M{1cm}|M{1cm}|M{1cm}|M{1cm}|M{1.3cm}|M{1.3cm}|} \hline 
时刻 & $0$ &&&&& $T$ \\ \hline 
时刻分划 & $t_0$ & $t_1$ & $t_2$ & $\cdots$ & $t_{n-1}$ & $t_n$ \\ \hline 
事件域 & $\mathcal{F}_{t_0}$ & $\mathcal{F}_{t_1}$ & $\mathcal{F}_{t_2}$ & $\cdots$ & $\mathcal{F}_{t_{n-1}}$ & $\mathcal{F}_{t_n}$ \\ \hline  
简单过程 & $Z_1$ & $Z_2$ & $Z_3$ & $\cdots$ & $Z_n$ &  \\ \hline
鞅差过程 &            & $\Delta_{1}B$ & $\Delta_{2}B$ & $\cdots$ & $\Delta_{n-1}B$ & $\Delta_{n}B$ \\ \hline
\end{tabular}
\end{table}

\item  当 $t_{k-1}\le t < t_k$ 时，
\begin{eqnarray*}
\int_0^t C_sdB_s = Z_1\Delta_1B + \cdots + Z_{k-1}\Delta_{k-1}B + Z_k(B_t - B_{t_{k-1}}). 
\end{eqnarray*}


\end{itemize}

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\begin{frame}{5.17. 证明从伊藤积分得到鞅 - 3 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
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\begin{itemize}%\itemsep1em

\item  {\color{red}“公平性质”：$\mathbb{E}(I_t(C)\mid \mathcal{F}_s) = I_s(C),\,\, s<t$. }

\item  证明：若 $s,t\in [t_{k-1},t_k]$. 根据简单过程的伊藤积分，有
\begin{eqnarray*}
I_t(C) &=& I_{t_{k-1}}(C) + Z_k(B_t-B_{t_{k-1}}) \\
&=& I_{t_{k-1}}(C) + Z_k(B_s-B_{t_{k-1}}) + Z_k(B_t-B_s) \\
&=& I_s(C) + Z_k(B_t-B_s). 
\end{eqnarray*}
根据 $Z_k$ 是 $\mathcal{F}_{t_{k-1}}$-可测，以及 $B_t-B_s$ 与 $\mathcal{F}_s$ 独立，得证。

\end{itemize}

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\begin{enumerate}\itemsep0.8em

\item  设 $B=\{B_t, t\in [0,T]\}$ 是一个布朗运动。
\item  设随机过程 $C$ 适应于这个布朗运动，即 $C_t$ 是 $\{B_s,s\in [0,t]\}$ 的函数。
\item  设积分 $\int_0^T \mathbb{E}(C_s^2)ds$ 是有限的。
\item  证明存在一列简单过程 $C^{(n)}$ 使得 $\int_0^T \mathbb{E} \left[ C-C^{(n)} \right]^2 ds\to 0. $
\item  证明存在随机过程 $I(C)$ 使得 $\mathbb{E} \sup\limits_{0\le t\le T} \left[ I_t(C) - I_t(C^{(n)}) \right]^2\to 0.$
\item  定义 $I_t(C) =\int_0^t C_sdB_s,\,\, t\in [0,T]$. 
\end{enumerate}

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\begin{frame}{5.19. Rule of thumb （经验法则）}

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\begin{itemize}\itemsep1em

\item  The Ito stochastic integral $I_t(C) = \int_0^t C_sdB_s,\, t\in [0,T]$, constitute a stochastic process. 
For a given partition $$\tau_n: 0 = t_0 < t_1 < t_2 < \cdots < t_n = T, $$
and $t\in [t_{k-1},t_k]$, the random variable $I_t(C)$ is 'close' to the Riemann-Stieltjes sum 
$$\sum\limits_{i=1}^{k-1} C_{t_{i-1}} (B_{t_i} - B_{t_{i-1}}) + C_{t_{k-1}}(B_t - B_{t_{k-1}}),$$
and this approximation is the closer (in the mean square sense) to the value of $I_t(C)$ the more dense the partition $\tau_n$ in $[0,T]$. 

\end{itemize}

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\begin{frame}{5.E.1. 单项选择 }

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\begin{itemize}\itemsep1em
\item  %第3题：单项选择
设随机变量序列 $A_n$ 均方收敛于随机变量 $A$. 下述说法中，正确的是哪个？

\begin{enumerate}
\item[a.]  随机变量 $A_n-A$ 的一阶矩 $\mathbb{E}(A_n-A)$ 收敛于零。
\item[b.]  随机变量 $A_n-A$ 的绝对值的一阶矩 $\mathbb{E}|A_n-A|$ 收敛于零。
\item[c.]  随机变量 $A_n-A$ 的二阶矩 $\mathbb{E}|A_n-A|^2$ 收敛于零。
\item[d.]  随机变量 $A_n$ 的二阶矩 $\mathbb{E}|A_n|^2$ 收敛于随机变量 $A$ 的二阶矩 $\mathbb{E}|A|^2$. 

\end{enumerate} 

\vspace{0.2cm}

\item  {\color{red}解答：c. 这个选项就是均方收敛的定义。 

}

\end{itemize}

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\begin{frame}{5.E.2. 多项选择 }

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\begin{itemize}
\item  %第4题：多项选择
设 $(\Omega, \mathcal{F}, P)$ 是一个概率空间。设 $\{B(t),t\ge 0\}$ 是标准布朗运动。设 $\tau_n:\,\,\, t_0<t_1<\cdots<t_n$ 是区间 $0\le t\le T$ 的一个分划。设 $$Q_n = \sum\limits_{k=1}^{n} \left[ B(t_k) - B(t_{k-1}) \right]^2.$$
下述说法中，正确是是哪些？
\begin{enumerate}
\item[a.]  当分划加细时，随机变量序列 $Q_n$ 均方收敛于常数 $T$. 
\item[b.]  当分划加细时，随机变量序列 $Q_n$ 绝对收敛于常数 $T$. 
\item[c.]  对每个分划，随机变量 $Q_n$ 服从正态分布。
\item[d.]  当分划加细时，随机变量序列 $Q_n$ 的方差收敛于零。

\end{enumerate} 

\vspace{0.2cm}

\item  {\color{red}解答：ad. 随机变量 $Q_n$ 的均值是区间的长度 $T$, 方差不超过 $2T\text{mesh}(\tau_n)$, 因此当分划加细时，随机变量序列 $Q_n$ 均方收敛于常数 $T$. 随机变量 $Q_n$ 是一些相互独立的正态分布的随机变量的平方和，因此不服从正态分布。

}

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\begin{itemize}
\item  %第5题：单项选择
设随机变量 $\xi_1, \xi_2, \xi_3$ 相互独立，且都服从标准正态分布。设 $\eta=\xi_1+2\xi_2+3\xi_3$. 求 $\eta$ 的分布。 
\begin{enumerate}
\item[a.]  卡方分布 $\chi^2(3)$. 
\item[b.]  卡方分布 $\chi^2(6)$. 
\item[c.]  正态分布 $N(0,6)$.
\item[d.]  正态分布 $N(0,14)$.
\end{enumerate} 

\vspace{0.2cm}

\item  {\color{red}解答：d. 根据正态分布的性质，$2\xi_2\sim N(0,4), 3\xi_3\sim N(0,9)$. 因为 $\xi_1, 2\xi_2, 3\xi_3$ 仍然相互独立，所以 $\eta\sim N(0,14)$. 

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\begin{itemize}
\item  %第6题：多项选择
设 $\{B(t),t\ge 0\}$ 是标准布朗运动。设 $C=\{C_t, t\in [0,3]\}$ 是如下定义的随机过程，其中 $Z_1,Z_2,Z_3$ 是三个随机变量。设下述伊藤积分存在。
\begin{eqnarray*}
C_t = \left\{ \begin{array}{ll}
Z_1, & 0\le t < 1, \\ 
Z_2, & 1\le t < 2, \\ 
Z_3, & 2\le t \le 3. 
\end{array}\right. \hspace{0.5cm} 
I_t(C) = \int_0^t C_sdB_s, \,\,\, 0\le t\le 3. 
\end{eqnarray*}
设 $\mathcal{F}_t = \sigma(B_s,0\le s\le t)$. 下述说法中，正确的是哪些？

\begin{enumerate}
\item[a.]  随机变量 $Z_0$ 是一个常数。
\item[b.]  随机变量 $Z_2$ 是 $\mathcal{F}_1$ 可测的。
\item[c.]  随机变量 $I_2(C)$ 等于 $Z_1B_1 + Z_2B_2$.
\item[d.]  随机变量 $I_3(C)$ 等于 $Z_1B_1 + Z_2(B_2-B_1) +Z_3(B_3-B_2)$. 
\end{enumerate} 

\vspace{0.2cm}

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\begin{itemize}

\item  {\color{red}解答：abd. 随机过程 $C$ 是一个简单过程。按照简单过程的伊藤积分的定义， 
\begin{eqnarray*}
I_1(C) &=& Z_1(B_1-B_0) = Z_1B_1, \\
I_2(C) &=& Z_1B_1 + Z_2(B_2-B_1), \\
I_3(C) &=& Z_1B_1 + Z_2(B_2-B_1) +Z_3(B_3-B_2). 
\end{eqnarray*}
因为伊藤积分的条件是简单过程 $C$ 是可预见的，而且 $\mathcal{F}_0=\{\varnothing,\Omega\}$, 所以选项 a 与 b 是对的。
选项 c 与伊藤积分的定义不符合，选项 d 则是符合的。

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\begin{itemize}
\item  %第7题：多项选择
在上一题中，我们有如下计算：
\begin{eqnarray*}
\mathbb{E}[I_3(C) \mid \mathcal{F}_2 ] %&=& \mathbb{E}[Z_1B_1 + Z_2(B_2-B_1)+Z_3(B_3-B_2) \mid \mathcal{F}_2 ] \\
&=&  \mathbb{E}[I_2(C) + Z_3(B_3-B_2) \mid \mathcal{F}_2 ] \\
&=&  \mathbb{E}[I_2(C) \mid \mathcal{F}_2 ] + \mathbb{E} [Z_3(B_3-B_2) \mid \mathcal{F}_2 ] \\
&=& I_2(C) + Z_3\mathbb{E} [(B_3-B_2) \mid \mathcal{F}_2 ] \\
&=& I_2(C) + Z_3\mathbb{E} (B_3-B_2) \\
&=& I_2(C).
\end{eqnarray*}
下述说法中，正确是是哪些？
\begin{enumerate}
\item[a.]  第一个等号的理由是简单过程的伊藤积分的定义。
\item[b.]  第三个等号的理由是随机变量 $I_2(C)$ 和随机变量 $Z_3$ 都是 $\mathcal{F}_2$ 可测的。
\item[c.]  第四个等号的理由是随机变量 $B_3-B_2$ 是 $\mathcal{F}_2$ 可测的。
\item[d.]  这个计算验证了伊藤积分的结果是一个鞅过程。

\end{enumerate} 

\vspace{0.2cm}

\item  {\color{red}解答：abd. 选项 c 不对。第四个等号的理由是随机变量 $B_3-B_2$ 与 $\mathcal{F}_2$ 是相互独立的。

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\begin{itemize}\itemsep1em
\item  %第8题：多项选择
设 $\{B(t),t\ge 0\}$ 是标准布朗运动。设 $f(x)=x^2$. 考虑下述计算：
\begin{eqnarray*}
B_t^2 = f(B_t) - f(0)  &=& \int_0^t f\,' (B_s)dB_s + \frac{1}{2} \int_0^t f\, ''(B_s)ds \\
&=& \int_0^t 2B_sdB_s + \frac{1}{2} \int_0^t 2ds 
=2\int_0^t B_sdB_s + t. 
\end{eqnarray*}
下述说法中，正确是是哪些？
\begin{enumerate}
\item[a.]  第二个等号的理由是伊藤引理。
\item[b.]  伊藤引理可以从泰勒公式出发形式地得到。
\item[c.]  第三个等号的理由是因为 $f(x)=x^2$ 所以有 $f\,'(x)=2x$ 以及 $f\,''(x)=2$. 
\item[d.]  标准布朗运动的定义条件之一是 $B_0=0$. 
\end{enumerate} 

%\vspace{0.2cm}

\item  {\color{red}解答：abcd. 这些说法都是对的。这是应用伊藤引理来计算伊藤积分 $\int_0^t B_sdB_s$ 的例子。

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\begin{itemize}
\item  %第9题：多项选择
设二元函数 $f(t,x)$ 有连续的二阶偏导数，一阶偏导数记为 $f_1$ 和 $f_2$, 二阶偏导数记为 $f_{11}, f_{12}$ 和 $f_{22}$. 
在 $f(t+dt,x+dx) - f(t,x)$ 的泰勒展开式包含下述哪些项？
\begin{enumerate}
\item[a.]  $f_1(t,x)dt$ 与 $f_2(t,x)dx$.  
\item[b.]  $f_{11}(t,x)(dt)^2$.
\item[c.]  $f_{12}(t,x)(dt)(dx)$.
\item[d.]  $f_{22}(t,x)(dx)^2$.
\end{enumerate} 

\vspace{0.2cm}

\item  {\color{red}解答：ac. 二元函数的泰勒公式为
\begin{eqnarray*}
f(t+dt,x+dx) - f(t,x)  &=& f_1(t,x)dt + f_2(t,x)dx + \frac{1}{2} f_{11}(t,x)(dt)^2 \\
&& + f_{12}(t,x)(dt)(dx) + \frac{1}{2}f_{22}(t,x)(dx)^2 + \cdots.
\end{eqnarray*}
因此选项 b 和 d 的系数应该是 $1/2$. 

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\begin{itemize}
\item  %第10题：多项选择
随机微分方程 $dX_t = a(t,X_t)dt +b(t,X_t)dB_t$ 的真实含义是下述等式，
$$X_t = X_0 + \int_0^t a(s,X_s)ds + \int_0^t b(s,X_s)ds, \,\,\, 0\le t\le T. $$
下述说法中，正确的是哪些？
\begin{enumerate}
\item[a.]  上式中的第一个积分是黎曼积分，第二个积分是伊藤积分。
\item[b.]  这个随机微分方程的未知函数是一个未知的随机过程。
\item[c.]  在系数 $a(t,x)$ 和 $b(t,x)$ 是两个线性函数的时候，这个方程存在唯一解。
\item[d.]  几何布朗运动是这个方程在 $a(t,x)=cx$ 与 $b(t,x)=\sigma x$ 的情形的解。

\end{enumerate} 

\vspace{0.2cm}

\item  {\color{red}解答：abcd. 这些说法都是对的。选项 c 是对的，因为线性函数关于变量 $x$ 满足 Lipschitz 条件。

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\begin{thebibliography}{99}

\bibitem{mikosch} {\ppr Thomas Mikosch}. {\ppr Elementary Stochastic Calculus}. 世界图书出版公司，{\ppr 2009} 年 {\ppr 8} 月第 {\ppr 1} 版。
\bibitem{wangjun} 王军、邵吉光、王娟. 随机过程及其在金融领域中的应用. 清华大学出版社，北京交通大学出版社，{\ppr 2018} 年{\ppr 8} 月第 {\ppr 2} 版。
\bibitem{zhangbo} 张波、商豪. 应用随机过程. 中国人民大学出版社，{\ppr 2016} 年 {\ppr 6} 月第 {\ppr 4} 版。
%\bibitem{karlin} {\ppr Mark A. Pinsky, Samuel Karlin}. {\ppr An Introduction to Stochastic Modeling}. 机械工业出版社，{\ppr 2013} 年 {\ppr 2} 月第 {\ppr 1} 版。
\bibitem{karatzas-shreve} {\ppr Ioannis Karatzas, Steven E. Shreve}. {\ppr Brownian Motion and Stochastic Calculus}. {\ppr GTM 113, Springer Verlag, New York, 1988. }



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